How do you find the derivative of #f(t)=t^(3/2)log_2sqrt(t+1)#?

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Answer 1 · 2017-12-23T05:50:42

#f'(t)=t^(3/2)(1/(2ln2(t+1)))+log_2sqrt(t+1)(3/2t^(1/2))#

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#f(t)=t^(3/2)log_2sqrt(t+1)#

This is a product of two functions. As such, we use the product rule which says if:

#f(x)=g(x)*h(x)#

#f'(x)=g(x)*h'(x)+h(x)*g'(x)#

In our problem,

#f(t)=g(t)*h(t)#

We need to figure out:

#f'(t)=g(t)*h'(t)+h(t)*g'(t)#

#g(t)=t^(3/2)#

#h(t)=log_2sqrt(t+1)#

#g'(t)=3/2t^(1/2)#

For #h'(t)#, we can say #u=sqrt(t+1)=(t+1)^(1/2)#

#du=1/2(t+1)^(-1/2)=1/(2(t+1)^(1/2))=1/(2sqrt(t+1)#

Then. we can say:

#h(t)=log_2u#

#h'(t)=(1/(u(ln2)))du#

#h'(t)=(1/(sqrt(t+1)(ln2)))(1/(2sqrt(t+1)))=1/(2ln2(t+1)#

Now, we can plug these results in:

#f'(t)=t^(3/2)(1/(2ln2(t+1)))+log_2sqrt(t+1)(3/2t^(1/2))#